Problems on Techniques of Integration

We have

\begin{displaymath}\frac{3 x^6 -2 x^5 + 32 x^4 -16 x^3 + 42x^2 + 18 x -27}{x^2 + 9} = 3x^4-2x^3+5x^2+2x-3 \;+\; \frac{x+1}{x^2 + 9}\cdot\end{displaymath}

So

\begin{displaymath}\int \frac{3 x^6 -2 x^5 + 32 x^4 -16 x^3 + 42x^2 + 18 x -27}{...
... + \frac{5x^3}{3}+ x^2 - 3x + \int \frac{x+1}{x^2 + 9} dx \cdot\end{displaymath}

In order to integrate $\displaystyle \frac{x+1}{x^2 + 9}$, we use the identity

\begin{displaymath}\frac{x+1}{x^2 + 9} = \frac{1}{2} \frac{2x}{(x^2 + 9)} + \frac{1}{x^2 + 9}\cdot\end{displaymath}

This implies

\begin{displaymath}\int \frac{x+1}{x^2 + 9}dx = \frac{1}{2} \int \frac{2x}{x^2 + 9}dx + \int \frac{1}{x^2 + 9}\cdot\end{displaymath}

On one hand, we have

\begin{displaymath}\int \frac{2x}{x^2 + 9}dx = \ln(x^2 + 9),\end{displaymath}

and also

\begin{displaymath}\int \frac{1}{x^2 + 9}dx = \frac{1}{3}\arctan\left(\frac{x}{3}\right).\end{displaymath}

We get

\begin{displaymath}\int \frac{x+1}{x^2 + 9}dx = \frac{1}{2} \ln(x^2 + 9) +\frac{1}{3}\arctan\left(\frac{x}{3}\right) + C.\end{displaymath}

Detailed Answer.


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